Question: Divide the following complex numbers. $ \dfrac{5+5i}{-1-2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1+2i}$ $ \dfrac{5+5i}{-1-2i} = \dfrac{5+5i}{-1-2i} \cdot \dfrac{{-1+2i}}{{-1+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(5+5i) \cdot (-1+2i)} {(-1-2i) \cdot (-1+2i)} = \dfrac{(5+5i) \cdot (-1+2i)} {(-1)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(5+5i) \cdot (-1+2i)} {(-1)^2 - (-2i)^2} = $ $ \dfrac{(5+5i) \cdot (-1+2i)} {1 + 4} = $ $ \dfrac{(5+5i) \cdot (-1+2i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({5+5i}) \cdot ({-1+2i})} {5} = $ $ \dfrac{{5} \cdot {(-1)} + {5} \cdot {(-1) i} + {5} \cdot {2 i} + {5} \cdot {2 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-5 - 5i + 10i + 10 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-5 - 5i + 10i - 10} {5} = \dfrac{-15 + 5i} {5} = -3+i $